All categories

3 years ago

(1.)What is the value of 35−3m , when m = 4?

1 answer:

4 0

(1.) 35-3m
m= 4

35-3m
= 35-3(4)
= 35-12 (do the multiple/division first before doing the addition/subtraction)
= 23

C. 23

(2.) 1 + x ÷ 5
x = 80

1 + x ÷ 5
= 1+80÷5
= 1+16
= 17

(3.) mx-y
m=5, x=3, and y=8

mx-y
= 5(3)-8
= 15-8
= 7

(4.) 3a+15+bc−6
a=7, b=3, and c=15

3a+15+bc-6
= 3(7)+15+3(15)-6
= 21+15+45-6
= 75

You might be interested in

If y = kx, what is the value of k if y = 20 and x = 0.4? A) 5 B) 8 C) 9.6 D) 50

Margaret [11]

Answer: D, 50
y = kx <span>y = 20 and x = 0.4
</span>
What you will do: You will plug in your y and your x.
Your equation should look like: 20=k(0.4)
Next: What you will do is divide each side of your equal sign by 0.4 so you will get your k alone.
You would get: k=50

HOPE THIS HELPS! ^_^

7 0

2 years ago

Help will give the branniest

Naily [24]

The answer is C... Hope it helps

4 0

3 years ago

Solve for X <br> 7x=49 (brainliest)

posledela

Answer:

$\huge\colorbox{violet}{ᴀɴsᴡᴇʀ}$

Hey There!

The answer = 7 = ( Seventh Track of TS8= Seven )

Step-by-step explanation:

$7x = 49 \\ x = 49 \div 7 \\ x = 7$

$7 \times 1 = 7 \\ 7 \times 2 = 14\\ 7 \times 3 = 21 \\ 7 \times 4 = 28 \\ 7 \times 5 = 35 \\ 7 \times 6 = 42 \\ \bold {7 \times 7 = 49} \\ 7 \times 8 = 56 \\ 7 \times 9 = 63 \\ 7 \times 10 = 7$

6 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

2 years ago

What is 1.25 times 10 to the 7th power in standard notation

Viefleur [7K]

Answer:

12,500,000. You simply move the decimal point over to the right 7 times.

5 0

3 years ago